BREAKING DOWN THE YEAR
This topic is a little math intensive and is not necessary for you to understand the rest of this chapter. The only sophisticated math, however, is handled using a spreadsheet function. If you’re willing to “go with” the use of the spreadsheet function and a brief explanation of what’ s happening, this section will give you a little more insight into the workings of Life Tables.
For whatever reason, an insurance company decided it would like to be able to sell term policies for half years rather than years. In order to price these policies, it
Table 10.6 Excerpt from the Life Table for Men with Some Curve Fitting Data

needed Life Tables with halfyear rather than fullyear steps. For a crude approximation, it could have just split the q yearly values in half, for example, for a man, q50 = 0.005648, so it could have used 0.005648/2 = 0.002824 for the first half of his fiftieth year and the same number for the second half. This is not a terrible approximation, but since the q values are climbing year by year in the 50year age range, the result would have been to overcharge for a first – half fiftieth year policy and to undercharge for a secondhalf fiftieth year policy.
My first step in creating a more accurate set of q values is to create a notation. The probability of a man dying sometime during his fiftieth year is q50. I’ll call the new set of numbers r to distinguish these numbers from the q numbers. Remember that there will be an r number for the first half of each year’s age and a different r number for the second half.
Table 10.6 starts with an excerpt from the men’s Life Table. I’m only considering ages 4554 and I’m only copying the age, q, and – columns. Remember that column l is the number of men alive out of the original 100,000. In the next column, I showed the number of men dead out of the original 100,000. This column was generated by simply subtracting the numbers in column l from 100,000. Figure 10.6 shows a plot of the number of men dead column versus the man’s age.
Spreadsheet programs offer the ability to “fit” a formula to a set of data. This means that the spreadsheet program takes a formula that you choose from a list of available formulas and adjusts the parameters in this formula to make a plot of it look as though this formula had actually created the original data. Some art is involved in the choice of formula. For the data shown in Figure 10.6, I chose a “second – order polynomial” as my formula. This is the part you don ’ t have to worry about if you’re uncomfortable with the math, but keep reading for the results.
The spreadsheet program calculated the formula
Number of dead = 18.09age2 – 1,304.1age + 27,906
5,000 45 47 49 51 53 55 Age (years) Figure 10.6 Curve fit used to generate a halfyear Life Table section. 
as its best fit to the data. In Table 10.6, I showed the formula predictions for the number of dead (labeled “by fit”) and the percent error in the formula prediction versus the original data. In Figure 10.6, I have superimposed a graph of the formula on top of the graph of the original data. As you can see from the table, the percent error is very small. And as you can see from the figure, it’s just about impossible to tell the difference between the two.
Table 10.7 shows how I built my new life table using my formula. The first column lists ages in halfyear steps. The second column lists the number of dead generated by using my formula. The third column lists the number of alive (,), generated by subtracting each number of dead value from the original 100,000. The fourth column is the Life Table d column, the number of men dying each year. This is generated by subtracting the number of alive in one row by the number of alive in the previous row, for example, d45.5 = l23 – 124.
Looking at the original Life Tables, d values are obtained by multiplying the number of people alive by the probability of their dying, dt = ql where i refers to any row (age) in the table. To get my new probability values, r, therefore, I just divide di by li. For example, r45.5 = d45.5/l45.5 = 176/93,980 = 0.001769.[28]
The insurance company would no doubt do much more exotic curve fitting than I did and extend its results over the entire 0 to 100year age span, but what I did above shows the calculations involved. I now have a Life Table that can accurately be used to prepare term policies in halfyear increments.
Table 10.7 Half Year Life Table for Men

1. A50year old man decides that he wants a $1,000,000 life insurance for 2 years.
(a) What will two separate 1year term policies cost, both bought at the beginning of the first year? Use 4% annual percentage rate (APR) for calculating present values and ignore insurance company markups.
(b) Repeat the above but treat this as a 2year policy bought on the man’s fiftieth birthday.
2. Suppose someone was to come down with an incurable terminal disease and was told by his or her doctor that “You have a twothird chance of lasting a year; you’ll never last more than 2 years.” Construct a Life Table for this person and calculate the cost of a $100,000 life insurance policy for this person.
3. Using the half year Life Table shown in Table 10.6, assuming an APR of 4% for the present value, what is the cost (idealized) of an 18month, $50,000 term policy for a man starting halfway through his forty – eighth year?
4. Life Tables can be generated for longer time steps than 1 year as well as for shorter time steps than 1 year. Tables generated in 5year steps are sometimes called abridged Life Tables. Starting with the women’s Life Table, generate an abridged Women’s Life Table.
5. You are taking a 5year business loan for $250,000 that you will amortize yearly with 60 equal monthly payments at an APR of 10%. Your lender wants a fully paidup life insurance policy to protect his or her interest. You will get a decreasing term policy that pays exactly the balance at the beginning of each year. You are a healthy, 35yearold male.
(a) Create an amortization table for the loan and show the balances at the beginning of each year.
(b) Using the balances at the beginning of each year from problem 5a, create the 5year decreasing term policy and get the price for this policy as an upfront payment. Use 5% as the APR for calculating present values. Assume that your insurance company prices this policy at 35% above the ideal calculation number.
(c) If you borrow the money for the policy along with your loan, what’s your effective APR (effective annual percentage rate [EAPR])?
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